# Where does this proof for arccosine derivative break down?

So I'm checking the identities for the derivative of the inverse cosine, and I ran against a wall:
With $\cos(y)=x$

$dy\over dx$$=$$1\over\sin(-y)$$=$$1\over-\sin(y)$

From here, I could either do $1\over\sqrt{1-\cos^2(-y)}$$=$$1\over\sqrt{1-\cos^2(y)}$ $=$$1\over\sqrt{1-x^2} Since cosine is an even function. Or I could go the simple way in which I get the correct answer: -1\over\sqrt{1-\cos^2(y)}$$=$$-1\over\sqrt{1-x^2} Why is my first answer incorrect? I suspect it was when I used the pythagorean identity, but I can't see why I'd get two different answers if I'm using identities only. Thanks in advance. • Why did you lose the minus sign on -\sin y in the first method when you converted to \sqrt{1 - \cos^2(-y)}? Also, why is it -y inside cosine and not just y? – tilper Aug 3 '16 at 1:39 • Since -sin(y)=sin(-y), instead of sticking with the negative sign outside, I placed it inside. Then, 1=cos^2(-y)+sin^2(-y). Or at least that was my reasoning. – take008 Aug 3 '16 at 1:44 ## 1 Answer The inverse cosine has domain [-1,1] and range [0,\pi], so your y is always between 0 and \pi. This means that \sin y\ge 0 and \sin(-y)\le 0. Thus, \sin(-1) must be the negative square root of 1-\cos^2(-y), i.e.,$$\sin(-y)=-\sqrt{1-\cos^2(-y)}=-\sqrt{1-\cos^2y}\;.$$Now the first method also gives you the correct result. • Oh, I get it. Basically I went out of the usual domain of definition when I placed the negative sign inside of$sin(-y)$, right? Which is where it led to funny business. – take008 Aug 3 '16 at 2:00 • @K.Takeuchi: There’s no problem with$\sin(-y)$; you just have to realize that$-y$has to be between$-\pi$and$0$, so that you need the negative square root of$1-\cos^2(-y)$. The real problem is that the identity$\sin^2x+\cos^2x=1$is not equivalent to$\sin x=\sqrt{1-\cos^2x}$:$\sin x$can be either$\sqrt{1-\cos^2x}$or$-\sqrt{1-\cos^2x}$, and which one it is depends on$x$. – Brian M. Scott Aug 3 '16 at 2:03 • Ok, I see. I'd never stopped to think about why on the formulas there's no$+-\$. I'll try doing some exercises to see if I can get a 100% understanding. – take008 Aug 3 '16 at 2:17
• @K.Takeuchi: Sounds like a good approach. I’m not really surprised that you hadn’t thought about it before: textbooks and teachers often slide over messy details involving signs of square roots, or just mention them in passing. – Brian M. Scott Aug 3 '16 at 2:20